Question: Solve for $X$. $X+\left[\begin{array}{rr}-12 & 15 \\ 15 & 6 \end{array}\right]=\left[\begin{array}{rr}2 & 8 \\ 1 & 3\end{array}\right] $ $X=$
Solution: The Strategy First, we can represent the matrices of the equation with letters, which will make the equation easier to handle. Then we can solve the equation for $X$ and obtain an expression with the letters we defined. Finally, we can substitute back the actual matrices into the resulting expression and simplify it. Solving the equation for $X$ We are given the following equation. $X+\left[\begin{array}{rr}-12 & 15 \\ 15 & 6 \end{array}\right]=\left[\begin{array}{rr}2 & 8 \\ 1 & 3\end{array}\right] $ Let's represent the above matrices as follows. $A=\left[\begin{array}{rr}-12 & 15 \\ 15 & 6 \end{array}\right] ~~~~~~~~~ B = \left[\begin{array}{rr}2 & 8 \\ 1 & 3\end{array}\right]$ Then we can rewrite the equation as follows. $X+A=B$ Now it's simple to solve the equation for $X$. $\begin{aligned}X+A&=B\\\\ X&=B-A \end{aligned}$ Finding $X$ We found that $X=B-A$. Now we can substitute the actual matrices back into the expression and simplify. $\begin{aligned}X&=B-A \\\\&=\left[\begin{array}{rr}2 & 8 \\ 1 & 3\end{array}\right]-\left[\begin{array}{rr}-12 & 15 \\ 15 & 6 \end{array}\right] \\\\\\&=\left[\begin{array}{rr}(2+12) & (8-15) \\ (1-15) & (3-6) \end{array}\right] \\\\\\&=\left[\begin{array}{rr}14 & -7 \\ -14 & -3\end{array}\right]\end{aligned}$ Summary $X=\left[\begin{array}{rr}14 & -7 \\ -14 & -3\end{array}\right]$